Not easy as sum of squares

Calculus Level 2

k = 1 1 k 2 \large \sum _{ k=1 }^{ \infty }{ \frac { 1 }{ k^{ 2 } } }

If the series above equals to π a b \frac{\pi^a}b for positive integers a a and b b , find the value of a + b a+b .


The answer is 8.

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1 solution

Shashwat Shukla
Jan 29, 2015

This is just ζ ( 2 ) = π 2 6 \zeta(2)=\frac{\pi^2}{6}

Thus a + b = 8 a+b=8 .

For proofs see: Basel Problem

As i dont know the answer to this problem so i m asking it here : Now if

S i = k = 1 i ( 36 k 2 1 ) i { S }_{ i }=\sum _{ k=1 }^{ \infty }{ \frac { i }{ { ({ 36k }^{ 2 }-1) }^{ i } } }

Find S 1 + S 2 { S }_{ 1 }+{ S }_{ 2 } ??????????????

(update) i have nailed it and posting it as a problem here

Gautam Sharma - 6 years, 4 months ago

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Same problem

U Z - 6 years, 4 months ago

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Yeah! Try the problem in my above comment. And pls explain.

Gautam Sharma - 6 years, 4 months ago

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